Show that (4-x)/(-1) = (y+3)/(-4) = (z+1)/7 and (x-1)/2 = (y+1)/(-3) = (z+10)/8 intersect each…

Line 1: $\frac{x-4}{1} = \frac{y+3}{-4} = \frac{z+1}{7} = \lambda$, so $(4 + \lambda,\, -3 - 4\lambda,\, -1 + 7\lambda)$. Line 2: $\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+10}{8} = \mu$, so $(1 + 2\mu,\, -1 - 3\mu,\, -10 + 8\mu)$. Equating $x$: $4 + \lambda = 1 + 2\mu \Rightarrow \lambda = 2\mu - 3$. Equating $y$: $-3 - 4\lambda = -1 - 3\mu$. Substituting $\lambda$: $9 - 8\mu = -1 - 3\mu \Rightarrow \mu = 2$, so $\lambda = 1$. Check $z$: $-1 + 7(1) = 6$ and $-10 + 8(2) = 6$. The values agree, so the lines intersect. Point of intersection: $(4 + 1,\, -3 - 4,\, -1 + 7) = (5, -7, 6)$.